Диссертация (1137342), страница 33

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Twist-field representations from twisted fermionsIt is easy to see that odd generators vanish U2k+1 = 0, while even generators in DNcase coincide with those in W (gl(N )) algebraU2k (z) =Dz2k−1NX: ψα∗ (z) · ψα (z) : + 21 Dz2k−1 Ψ(z) · Ψ(z),k = 1, 2, . . .(6.26)α=1So finally we have the following sets of independent generators:U1 (z), U2 (z), . .

. , UN (z) for W (gl(N ))U2 (z), U4 (z), . . . , U2N −2 (z), V (z) for W (so(2N ))U2 (z), U4 (z), . . . , U2N (z), V (z) for W (so(2N + 1))(6.27)Twist-field representations from twisted fermionsFermions and W-algebrasFor any current algebra, generated by currents {AI (z)}, the commutation relationsfollow from their local OPE’sX (AI AJ )K (w)AI (z)AJ (w) =(6.28)z→w(z − w)KKHowever, to define the commutation relations in addition to local OPE’s one shouldalso know the boundary conditions for the currents: in radial quantization – theanalytic behaviour of AI (z) around zero.

Any vertex operator Vg (0), e.g. sitting atthe origin 3 , can create nontrivial monodromy for our currents:XAI (e2πi z)Vg (0) =gIJ AJ (z)Vg (0)(6.29)jfor some linear automorphism of the current algebra.Perhaps the simplest example of such nontrivial monodromy is the diagonal transformation of the fermionic fieldsψα∗ (e2πi z) = e2πiθα ψα∗ (z),ψα (e2πi z) = e−2πiθα ψα (z),α = 1, .

. . , N(6.30)which just shifts the mode expansion indexψα∗ (z)=X∗ψα,p1p∈Z+ 21z p+ 2 −θα,ψα (z) =X∗ψα,p1p∈Z+ 21z p+ 2 +θα(6.31)Instead of the OPE (6.2) one gets thereforez θα w−θβ+ z θα w−θβ : ψα∗ (z)ψβ (w) : =z−w(6.32)1θα δαβ∗++ : ψα (w)ψβ (w) : +reg.z−wwψα∗ (z)ψβ (w) → z θα w−θβ ψα∗ (z)ψβ (w) ==z→w3For nontrivial boundary conditions we assume presence of such field by default, when obvious –not indicating it explicitly.1576. Twist-field representations of W-algebras, exact conformal blocks and character identitieswhich means that for the shifted fermions (6.31) one should use different normalordering:θα δαβ(6.33)(ψα∗ (z)ψβ (z)) =+ : ψα∗ (z)ψβ (z) :z\) algebra one has extra shiftThis implies that for the diagonal components gl(N1σαJα (z) → Jα (z) + z , while for the non-diagonal currents we obtainJαβ (z) =Xn∈ZJαβ,nn+1+θα −θβz(6.34)so that the commutation relations for this “twisted” Kac-Moody algebra become[Jαβ,n , Jγδ,m ] = (n − θα + θβ )δn+m,0 δβγ δαδ + δβγ Jαδ,m+n − δαδ Jβγ,m+n(6.35)We see that these commutation relations differ from the conventional ones (6.7) onlyby the extra shift which can be hidden into the Cartan generators Jαα,0 .

However,\) does not contain zero modes, and we cannot think aboutin the twisted case gl(N1the W-algebra as about commutant of some gl(N ). But nevertheless we define thecurrentsNXk−1(6.36)Uk (z) = Dz(ψα∗ (z) · ψα (z))α=1One can still use two basic facts:• since Uk (e2πi z) = Uk (z), they are expanded in integer powers of z as before;• they satisfy the same algebraic relations for all values of monodromies {σα },because the OPE’s of ψα , ψα∗ (and so the OPE’s of Uk ) do not depend on thesemonodromy parameters.Twist fields and Cartan’s normalizersConsider now more general situation, whenψα∗ (e2πi z) =NXgαβ ψβ∗ (z),ψα (e2πi z) =β=1NX−1gβαψβ (z)(6.37)β=1i.e. compare to (6.30) the monodromy is no longer diagonal 4 . It is clear that then\) isthe action on gl(N1Jαβ (z) 7→ gαα0 gβ−1(6.38)0 β Jα0 β 0 (z)The most general transformation we consider in the O(n) case mixes ψ and ψ ∗ :ψα (e2πiz) =NXgαβ ψβ (z),α = −N, . .

. , N(6.39)β=−N4This element should preserve the structure of the OPEs, so it should preserve symmetric form onfermions, and lies therefore in O(2N ). Notice that it automatically implies that all even generatorsof the W-algebra U2k (w) are also preserved. To preserve odd generators U2k+1 (z) one should havealso g ∈ Sp(2N ), but O(2N ) ∩ Sp(2N ) = GL(N ), so g ∈ GL(N ).1586.4. Twist-field representations from twisted fermions∗= ψα , α > 0, and ψ0 can bewhere it is convenient to introduce conventions ψ−αabsent.

Matrix g here should preserve the anticommutation relations.Definition 1. We call the vertex operator Vg = Og a twist field when g lies in thenormalizer of Cartan h ⊂ g, i.e. g ∈ NG (h) iffghg −1 = h(6.40)Such elements also preserve the Heisenberg subalgebra bh = hJ1 (z), . .

. , Jrank g (z)i ⊂bg1gbhg −1 = bh(6.41)We are going now to discuss the structure of the Cartan normalizers NGL(N ) (h) andNO(n) (h), which classify the twist fields for the WN = W (gl(N )) and W (so(n)) (foreven n = 2N and odd n = 2N + 1) correspondingly.Structure of the Cartan normalizer for gl(N ). Let us choose the Cartan subalgebra in a standard way h ⊃ diag(x1 , . . . , xN ), so conjugation (6.40) can only permutethe eigenvalues. Therefore we conclude thatN(6.42)g = s · (λ1 , . .

. , λN ) ∈ SN n C× = NGL(N ) (h)or just g is a quasipermutation.Let us now find the conjugacy classes in this group. Any element of NGL(N ) (h)has the form g = (c1 . . . ck , (λ1 , . . . , λN )), where ci are the cyclic permutations – theironly parameters are lengths lj = l(cj ). It is enough to consider just a single cycle ofthe length l = l(c)g = (c, (λ1 , . . . , λl ))(6.43)since any g can be decomposed into a product of such elements. Conjugation of thiselement by diagonal matrix is given by(1, (µ1 , . . .

, µl )) · (c, (λ1 , . . . , λl )) · (1, (µ1 , . . . , µl ))−1 =µ2µlµ1= (c, (λ1 , λ2 , . . . , λl ))µ2µ3µ1(6.44)Therefore one can always adjust {µi } to get rid of all {λi } except for one, e.g. to putQ1/lλi 7→ λ̄ = li=1 λi = e2πir , these “averaged over a cycle” parameters have been calledas r-charges in [GMtw]. Hence, all elements of g ∈ NGL(N ) (h) can be conjugated tothe products over the cycles[g] ∼KKYY[lj , λ̄j ] =[lj , e2πirj ]j=1(6.45)j=1Structure of NO(n) (h).

Using complexification of fermions (6.20) we rewrite thenNNPPPquadratic form ds2 =dΨ2i as ds2 =dψα∗ dψα + dΨ2 =dψ−α dψα + dΨ2 (thei=1α=1α=1last term is present only for the BN -series). In this basis the so(n) algebra (the algebra,preserving this form) becomes just the algebra of matrices, which are antisymmetric1596. Twist-field representations of W-algebras, exact conformal blocks and character identitiesunder the reflection w.r.t. the anti-diagonal. In particular, the Cartan elements aregiven byh 3 diag(x1 , . .

. , xN , 0, −xN , . . . , −x1 )(6.46)for BN -series (and for the DN -series 0 in the middle just should be removed). Theaction of an element from NO(n) (h) should preserve the chosen quadratic form, and,when acting on the diagonal matrix (6.46), it can only permute some eigenvalues, alsodoing it simultaneously in the both blocks, or interchange xα with −xα (the same asto change the sign of xα ). It is defined in this way up to a subgroup of diagonalmatrices themselves. In other wordsNO(2N ) (h) = SN n (Z/2Z)N n (C× )NNO(2N +1) (h) = NO(2N ) (h) × Z/2Z(6.47)where the last factor Z/2Z comes from changing sign of the extra fermion Ψ.

Thistriple (s, ~σ , ~λ) ∈ NO(n) (h), with s ∈ SN , σα ∈ Z/2Z and λα ∈ C× , is embedded intoO(n) as follows∗SN : ({α 7→ s(α)}, 1, 1) = {ψα 7→ ψs(α) ; ψα∗ 7→ ψs(α)}(Z/2Z)N : (1, ~σ , 1) = {ψα →7 ψσα α }× N∗(C ) : (1, 1, ~λ) = {ψα 7→ λα ψα ; ψα∗ 7→ λ−1α ψα }(6.48)∗and in these formulas ψ−α = ψα∗ and ψ−α= ψα is again implied. The structure of theactions in the semidirect product has the obvious from:~σ : λα 7→ λσαα ,s : (σα , λα ) 7→ (σs(α) , λs(α) )(6.49)Notice that normalizer of Cartan in SO(n)NSO(n) (h) = SO(n) ∩ NO(n) (h)(6.50)Qis distinguished by condition that Nα=1 σα = 1, and the Weyl group is given as thefactor of this normalizer by the Cartan torusW(so(n)) = NSO(n) (h)/H(6.51)Consider now the conjugacy classes in NO(n) (h).

First, conjugating an arbitrary element (s, ~σ , ~λ) by permutations, we again reduce the problem to the case when s = cis just a single cycle. Then one can further conjugate this element by (Z/2Z)N :(1,~, 1) · (c, ~σ , •) · (1,~, 1)−1 7→ (c, (σ1 · 1 2 , σ2 2 3 , . . . , σN · 1 N ), •)(6.52)and solving equations for {α } remove all σα = −1) except for, maybe, one.

Hence:• For σ = (1, . . . , 1) the situation is the same as in gl(N ) case: we can transform~λ to (λ, . . . , λ). These conjugacy classes are therefore (denoted by [l, λ]+ )Y 1/l(c)(c, 1, ~λ) ∼ [l(c),λi]+(6.53)1606.4. Twist-field representations from twisted fermions• For, say, σ = (−1, 1, . . . , 1) let us conjugate this element by (1, 1, µ~ ):(1, 1, µ~ )(c, (−1, 1, . . .

1), ~λ)(1, 1, µ~ )−1 = (c, (−1, 1, . . . , 1), ~λ0 )~λ0 = (λ1 µ1 µ−1 , λ2 µ2 µ−1 , . . . , λl−1 µl−1 µ−1 , λl µl µ1 )32(6.54)lIn contrast to Qthe previous case, here one can put all λ0i = 1, since one canput first µ21 = i λ−1i , and then solve N − 1 equations µi+1 = λi µi not beingrestricted by any boundary conditions. It means that(c, (−1, 1, . . . , 1), ~λ) ∼ [l(c)]−(6.55)Therefore we can formulate:Lemma 6.1. One gets for the conjugacy classes0KKYYNO(2N ) (h) : g ∼[lj , λj ]+ ·[lj ]−j=1NO(2N +1) (h) : g ∼ [] ·j=1KY(6.56)0KY[lj , λj ]+ ·[lj ]−j=1j=1and we are now ready to describe the twist fields in detail.Corollary: Formulas (6.45) and (6.56) give also classification of the conjugacy classesin the Weyl groups W(gl(N )) = SN and W(so(n)).

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